BE Civil Engineering (IOE, TU) Engineering Mathematics I (IOE, SH 401) Question Paper 2078 Nepal

Leibnitz's Theorem. If $u$ and $v$ are functions of $x$ that are each differentiable $n$ times, then the $n^{th}$ derivative of their product is

$$\frac{d^n}{dx^n}(uv) = \sum_{r=0}^{n}\binom{n}{r}\,u_{n-r}\,v_r = u_n v + \binom{n}{1}u_{n-1}v_1 + \binom{n}{2}u_{n-2}v_2 + \cdots + u\,v_n,$$

where $u_k = \dfrac{d^k u}{dx^k}$ and $v_k = \dfrac{d^k v}{dx^k}$.

Finding the $n^{th}$ derivative of $y = e^{2x}\sin 3x$.

For a function of the form $y = e^{ax}\sin(bx)$ there is a standard result. Differentiate once:

$$y_1 = 2e^{2x}\sin 3x + 3e^{2x}\cos 3x = e^{2x}(2\sin 3x + 3\cos 3x).$$

Write $2\sin 3x + 3\cos 3x$ in the form $R\sin(3x+\phi)$ with $R = \sqrt{2^2+3^2} = \sqrt{13}$ and $\tan\phi = \dfrac{3}{2}$. Then

$$y_1 = \sqrt{13}\,e^{2x}\sin\!\left(3x+\phi\right),\qquad \phi = \tan^{-1}\tfrac{3}{2}.$$

Each differentiation multiplies by $\sqrt{13}$ and advances the phase by $\phi$. Hence, by induction,

$$\boxed{\,y_n = \left(\sqrt{13}\right)^{n} e^{2x}\,\sin\!\left(3x + n\tan^{-1}\tfrac{3}{2}\right) = 13^{\,n/2}\,e^{2x}\,\sin\!\left(3x + n\tan^{-1}\tfrac{3}{2}\right).}$$

Verification for $n = 2$ (direct). Differentiate $y_1 = e^{2x}(2\sin 3x + 3\cos 3x)$:

$$y_2 = 2e^{2x}(2\sin 3x + 3\cos 3x) + e^{2x}(6\cos 3x - 9\sin 3x)$$ $$= e^{2x}\big[(4-9)\sin 3x + (6+6)\cos 3x\big] = e^{2x}(-5\sin 3x + 12\cos 3x).$$

Verification from the general formula ($n=2$). Here $R^2 = 13$, $\cos\phi = 2/\sqrt{13}$, $\sin\phi = 3/\sqrt{13}$, so $\cos 2\phi = 1 - 2\sin^2\phi = 1 - 2\cdot\frac{9}{13} = -\frac{5}{13}$ and $\sin 2\phi = 2\sin\phi\cos\phi = 2\cdot\frac{3}{\sqrt{13}}\cdot\frac{2}{\sqrt{13}} = \frac{12}{13}$.

$$y_2 = 13\,e^{2x}\sin(3x+2\phi) = 13\,e^{2x}\big[\sin 3x\cos 2\phi + \cos 3x\sin 2\phi\big] = 13\,e^{2x}\Big[\sin 3x\!\left(-\tfrac{5}{13}\right) + \cos 3x\!\left(\tfrac{12}{13}\right)\Big]$$ $$= e^{2x}(-5\sin 3x + 12\cos 3x).$$

Both methods agree, confirming the general formula. $y_n = 13^{n/2}e^{2x}\sin\!\big(3x + n\tan^{-1}\tfrac32\big)$.

Euler's theorem. If $f(x,y)$ is a homogeneous function of degree $n$, i.e. $f(tx,ty)=t^n f(x,y)$, then

$$x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = n\,f.$$

Setting up. Let $u = \tan^{-1} z$ where $z = \dfrac{x^3+y^3}{x-y}$. Note $u$ itself is not homogeneous, but $z = \tan u$ is. Check the degree of $z$:

$$z(tx,ty) = \frac{t^3(x^3+y^3)}{t(x-y)} = t^2\,\frac{x^3+y^3}{x-y} = t^2 z.$$

So $z = \tan u$ is homogeneous of degree $n = 2$. Apply Euler's theorem to $z$:

$$x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = 2z. \tag{1}$$

Converting derivatives of $z$ to derivatives of $u$. Since $z = \tan u$,

$$\frac{\partial z}{\partial x} = \sec^2 u\,\frac{\partial u}{\partial x}, \qquad \frac{\partial z}{\partial y} = \sec^2 u\,\frac{\partial u}{\partial y}.$$

Substitute into (1):

$$x\sec^2 u\,\frac{\partial u}{\partial x} + y\sec^2 u\,\frac{\partial u}{\partial y} = 2\tan u.$$

Divide both sides by $\sec^2 u$:

$$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac{2\tan u}{\sec^2 u} = 2\,\frac{\sin u}{\cos u}\cdot\cos^2 u = 2\sin u\cos u = \sin 2u.$$

Therefore

$$\boxed{\,x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \sin 2u.\,}$$

This is the required result.

Deriving the reduction formula. Write $\sin^n x = \sin^{n-1}x\cdot\sin x$ and integrate by parts over $[0,\pi/2]$ with $u=\sin^{n-1}x$, $dv=\sin x\,dx$, so $du=(n-1)\sin^{n-2}x\cos x\,dx$ and $v=-\cos x$:

$$I_n = \Big[-\sin^{n-1}x\cos x\Big]_0^{\pi/2} + (n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2 x\,dx.$$

The boundary term vanishes (at $x=\pi/2$, $\cos x=0$; at $x=0$, $\sin x=0$ for $n\ge 2$). Using $\cos^2 x = 1-\sin^2 x$:

$$I_n = (n-1)\int_0^{\pi/2}\sin^{n-2}x\,dx - (n-1)\int_0^{\pi/2}\sin^{n}x\,dx = (n-1)I_{n-2} - (n-1)I_n.$$

Collecting $I_n$ terms: $I_n + (n-1)I_n = (n-1)I_{n-2}$, i.e. $nI_n = (n-1)I_{n-2}$, giving the reduction formula

$$\boxed{\,I_n = \frac{n-1}{n}\,I_{n-2},\quad n\ge 2.\,}$$

Base value. $I_0 = \displaystyle\int_0^{\pi/2}dx = \frac{\pi}{2}.$

Evaluating $\int_0^{\pi/2}\sin^6 x\,dx = I_6$. Apply the formula repeatedly:

$$I_6 = \frac{5}{6}I_4 = \frac{5}{6}\cdot\frac{3}{4}I_2 = \frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}I_0 = \frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}.$$

Multiply the fractions: $\dfrac{5\cdot 3\cdot 1}{6\cdot 4\cdot 2} = \dfrac{15}{48} = \dfrac{5}{16}$. Hence

$$I_6 = \frac{5}{16}\cdot\frac{\pi}{2} = \frac{5\pi}{32}.$$ $$\boxed{\,\int_0^{\pi/2}\sin^6 x\,dx = \dfrac{5\pi}{32} \approx 0.4909.\,}$$

Group and complete the square.

$$4x^2 - 16x + 9y^2 + 18y - 11 = 0.$$ $$4(x^2 - 4x) + 9(y^2 + 2y) = 11.$$

Complete the square inside each bracket: $x^2-4x = (x-2)^2 - 4$ and $y^2+2y = (y+1)^2 - 1$.

$$4\big[(x-2)^2 - 4\big] + 9\big[(y+1)^2 - 1\big] = 11$$ $$4(x-2)^2 - 16 + 9(y+1)^2 - 9 = 11$$ $$4(x-2)^2 + 9(y+1)^2 = 36.$$

Divide through by $36$:

$$\boxed{\,\frac{(x-2)^2}{9} + \frac{(y+1)^2}{4} = 1.\,}$$

Identification. Both squared terms are positive with unequal denominators, so this is an ellipse with major axis parallel to the $x$-axis (since $9 > 4$).

Centre. $(h,k) = (2,\,-1)$.

Semi-axes. $a^2 = 9 \Rightarrow a = 3$ (semi-major), $b^2 = 4 \Rightarrow b = 2$ (semi-minor).

• Length of major axis $= 2a = 6$ units.
• Length of minor axis $= 2b = 4$ units.

Eccentricity. For an ellipse, $b^2 = a^2(1-e^2)$, so

$$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \approx 0.745.$$

Foci. Distance from centre to focus is $ae = 3\cdot\dfrac{\sqrt5}{3} = \sqrt5$. Since the major axis is horizontal, the foci lie at $(h\pm ae,\,k)$:

$$\boxed{\,\big(2+\sqrt5,\,-1\big)\ \text{and}\ \big(2-\sqrt5,\,-1\big)\ \approx\ (4.236,\,-1)\ \text{and}\ (-0.236,\,-1).}$$

Summary table.

Recognise the standard linear form. The equation is $\dfrac{dy}{dx} + P(x)\,y = Q(x)$ with

$$P(x) = \frac{2}{x}, \qquad Q(x) = 6x^2.$$

Integrating factor (IF).

$$\mu(x) = e^{\int P\,dx} = e^{\int \frac{2}{x}\,dx} = e^{2\ln x} = e^{\ln x^2} = x^2.$$

Multiply through by the IF. Multiplying the ODE by $x^2$:

$$x^2\frac{dy}{dx} + 2x\,y = 6x^4.$$

The left-hand side is exactly $\dfrac{d}{dx}\big(x^2 y\big)$:

$$\frac{d}{dx}\big(x^2 y\big) = 6x^4.$$

Integrate both sides with respect to $x$.

$$x^2 y = \int 6x^4\,dx = \frac{6x^5}{5} + C.$$

General solution. Divide by $x^2$:

$$\boxed{\,y = \frac{6x^3}{5} + \frac{C}{x^2}.\,}$$

Apply the initial condition $y(1)=4$.

$$4 = \frac{6(1)^3}{5} + \frac{C}{1^2} = \frac{6}{5} + C \;\Rightarrow\; C = 4 - \frac{6}{5} = \frac{20-6}{5} = \frac{14}{5}.$$

Particular solution.

$$\boxed{\,y = \frac{6x^3}{5} + \frac{14}{5x^2} = \frac{6x^5 + 14}{5x^2}.\,}$$

Check at $x=1$: $y = \dfrac{6+14}{5} = \dfrac{20}{5} = 4.$ ✓

Point on the curve. At $x=2$:

$$y = 2^3 - 3(2)^2 + 2 = 8 - 12 + 2 = -2.$$

So the point is $P(2,\,-2)$.

Slope of the tangent. Differentiate:

$$\frac{dy}{dx} = 3x^2 - 6x.$$

At $x=2$: $\;m = 3(4) - 6(2) = 12 - 12 = 0.$

The tangent slope is $m = 0$, so the tangent is horizontal.

Tangent line. Using $y - y_1 = m(x - x_1)$:

$$y - (-2) = 0\,(x - 2) \;\Rightarrow\; \boxed{\,y = -2.\,}$$

Normal line. The normal is perpendicular to the tangent. Since the tangent is horizontal, the normal is vertical (its slope is undefined). It passes through $x=2$:

$$\boxed{\,x = 2.\,}$$

Summary: Tangent: $y=-2$ (horizontal); Normal: $x=2$ (vertical). This point $(2,-2)$ is in fact a local minimum of the cubic, consistent with the horizontal tangent.

Set up variables. Let the side parallel to the river be $x$ (m) and the two sides perpendicular to the river be $y$ (m) each. Only three sides are fenced (the river side is free).

Constraint (fixed area):

$$x\,y = 1800 \;\Rightarrow\; x = \frac{1800}{y}.$$

Objective (fencing length to minimise):

$$L = x + 2y.$$

Substitute the constraint:

$$L(y) = \frac{1800}{y} + 2y.$$

Minimise. Differentiate and set to zero:

$$\frac{dL}{dy} = -\frac{1800}{y^2} + 2 = 0 \;\Rightarrow\; 2 = \frac{1800}{y^2} \;\Rightarrow\; y^2 = 900 \;\Rightarrow\; y = 30\ \text{m}.$$

(Take the positive root since $y>0$.)

Second-derivative test.

$$\frac{d^2L}{dy^2} = \frac{3600}{y^3} > 0 \quad\text{for } y>0,$$

so $y=30$ m gives a minimum.

Find $x$.

$$x = \frac{1800}{30} = 60\ \text{m}.$$

Minimum fencing length.

$$L = 60 + 2(30) = 60 + 60 = 120\ \text{m}.$$ $$\boxed{\text{Dimensions: } 60\ \text{m (along river)} \times 30\ \text{m; minimum fencing} = 120\ \text{m}.}$$

Chain rule for $\partial z/\partial r$. Since $z$ depends on $x$ and $y$, which depend on $r$ (and $\theta$):

$$\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}.$$

With $x = r\cos\theta$, $y = r\sin\theta$ we have $\dfrac{\partial x}{\partial r} = \cos\theta$ and $\dfrac{\partial y}{\partial r} = \sin\theta$. Hence

$$\boxed{\,\frac{\partial z}{\partial r} = \cos\theta\,\frac{\partial z}{\partial x} + \sin\theta\,\frac{\partial z}{\partial y}.\,}$$

Verification for $z = x^2 + y^2$.

Right-hand side (via chain rule). $\dfrac{\partial z}{\partial x} = 2x$, $\dfrac{\partial z}{\partial y} = 2y$, so

$$\frac{\partial z}{\partial r} = \cos\theta\,(2x) + \sin\theta\,(2y) = 2(r\cos\theta)\cos\theta + 2(r\sin\theta)\sin\theta$$ $$= 2r(\cos^2\theta + \sin^2\theta) = 2r.$$

Left-hand side (direct substitution). In polar form,

$$z = x^2 + y^2 = r^2\cos^2\theta + r^2\sin^2\theta = r^2,$$

so

$$\frac{\partial z}{\partial r} = \frac{\partial}{\partial r}(r^2) = 2r.$$

Both give $\dfrac{\partial z}{\partial r} = 2r$, confirming the chain-rule result. Verified.

Find the points of intersection. Set $x^2 = x + 2$:

$$x^2 - x - 2 = 0 \;\Rightarrow\; (x-2)(x+1) = 0 \;\Rightarrow\; x = -1\ \text{or}\ x = 2.$$

So the curves meet at $x = -1$ and $x = 2$.

Identify the upper and lower curves. On the interval $(-1,2)$, test $x=0$: line gives $y = 0+2 = 2$, parabola gives $y = 0$. The line lies above the parabola throughout the interval.

Set up the area integral.

$$A = \int_{-1}^{2}\big[(x+2) - x^2\big]\,dx.$$

Integrate.

$$A = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}.$$

At $x = 2$: $\dfrac{4}{2} + 4 - \dfrac{8}{3} = 2 + 4 - \dfrac{8}{3} = 6 - \dfrac{8}{3} = \dfrac{18-8}{3} = \dfrac{10}{3}.$

At $x = -1$: $\dfrac{1}{2} - 2 - \dfrac{-1}{3} = \dfrac{1}{2} - 2 + \dfrac{1}{3} = \dfrac{3 - 12 + 2}{6} = \dfrac{-7}{6}.$

Subtract.

$$A = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}.$$ $$\boxed{\,A = \dfrac{9}{2} = 4.5\ \text{square units}.\,}$$

Perpendicular distance formula. The distance from a point $(x_0,y_0)$ to the line $Ax + By + C = 0$ is

$$d = \frac{|A x_0 + B y_0 + C|}{\sqrt{A^2 + B^2}}.$$

Here $A = 5$, $B = -12$, $C = 2$, and $(x_0,y_0) = (3,-4)$.

Numerator.

$$A x_0 + B y_0 + C = 5(3) + (-12)(-4) + 2 = 15 + 48 + 2 = 65.$$

So $|{\cdots}| = 65$.

Denominator.

$$\sqrt{A^2 + B^2} = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13.$$

Distance.

$$d = \frac{65}{13} = 5.$$ $$\boxed{\,d = 5\ \text{units}.\,}$$

Line through $(3,-4)$ parallel to the given line. A parallel line has the same coefficients of $x$ and $y$: $5x - 12y + k = 0$. Substitute $(3,-4)$:

$$5(3) - 12(-4) + k = 0 \;\Rightarrow\; 15 + 48 + k = 0 \;\Rightarrow\; k = -63.$$ $$\boxed{\,5x - 12y - 63 = 0.\,}$$

Recognise the type. Writing it as $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{2xy}$, the right side is a function of $y/x$ only — this is a homogeneous first-order ODE.

Substitute $y = vx$, so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$. Then

$$v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{2x\cdot vx} = \frac{x^2(1 + v^2)}{2vx^2} = \frac{1 + v^2}{2v}.$$

Isolate the derivative term.

$$x\frac{dv}{dx} = \frac{1+v^2}{2v} - v = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v}.$$

Separate variables.

$$\frac{2v}{1 - v^2}\,dv = \frac{dx}{x}.$$

Integrate. For the left side, note that $\dfrac{d}{dv}(1 - v^2) = -2v$, so $\displaystyle\int \frac{2v}{1-v^2}\,dv = -\ln|1 - v^2|$. Thus

$$-\ln|1 - v^2| = \ln|x| + C_1.$$

Combine logarithms. Write $C_1 = -\ln|C|$ for convenience:

$$-\ln|1 - v^2| = \ln|x| - \ln|C| \;\Rightarrow\; \ln\frac{1}{|1-v^2|} = \ln\frac{|x|}{|C|}.$$

Exponentiate: $\dfrac{1}{1 - v^2} = \dfrac{x}{C}$, i.e. $C = x(1 - v^2)$.

Back-substitute $v = y/x$, so $1 - v^2 = 1 - \dfrac{y^2}{x^2} = \dfrac{x^2 - y^2}{x^2}$:

$$C = x\cdot\frac{x^2 - y^2}{x^2} = \frac{x^2 - y^2}{x}.$$

General solution.

$$\boxed{\,x^2 - y^2 = Cx,\,}$$

where $C$ is an arbitrary constant.

Check. Differentiate $x^2 - y^2 = Cx$ implicitly: $2x - 2y\,y' = C$. From the solution $C = (x^2-y^2)/x$, so $2x - 2y\,y' = (x^2-y^2)/x$. Multiply by $x$: $2x^2 - 2xy\,y' = x^2 - y^2$, giving $2xy\,y' = 2x^2 - x^2 + y^2 = x^2 + y^2$, i.e. $\dfrac{dy}{dx} = \dfrac{x^2+y^2}{2xy}$. ✓ This matches the original equation.